Qubit Physics

In practice, to correctly perform quantization via the commutator relation, it is essential to strictly define the correspondence between \(x, p\); otherwise, sign discrepancies might arise. Continuing from the previous section, we know: \[ \dot{\Phi}_\# = \frac{d}{dt} \frac{\partial \mathcal{L}}{\partial \dot{Q}} = \frac{\partial \mathcal{L}}{\partial Q} = -\frac{Q}{C} \implies Q = -C\dot{\Phi}_\#. \] Now, we transform the Lagrangian \(L\) into momentum space.

Define a new Lagrangian \(\bar{\mathcal{L}}\) as: \[ \bar{\mathcal{L}} \equiv \mathcal{L} - \frac{d}{dt} (\Phi_\# Q) = \mathcal{L} - \dot{\Phi}_\# Q - \Phi_\# \dot{Q}, \] Expanding this yields: \[ \bar{\mathcal{L}} = \frac{L\dot{Q}^2}{2} - \frac{Q^2}{2C} - \dot{\Phi}_\# Q - \Phi_\# \dot{Q} = \frac{\Phi_\#^2}{2L} - \frac{C\dot{\Phi}_\#^2}{2} + C\dot{\Phi}_\#^2 - \frac{\Phi_\#^2}{L}, \] Simplifying to: \[ \bar{\mathcal{L}} = \frac{C\dot{\Phi}_\#^2}{2} - \frac{\Phi_\#^2}{2L}. \]

Note that at this point \(\mathcal{L}(Q, \dot{Q}) \to \bar{\mathcal{L}}(\Phi_\#, \dot{\Phi}_\#)\), and the roles of kinetic and potential energy are interchanged. Additionally, \(\bar{\mathcal{L}} = -\mathcal{L}\), corresponding to the antisymmetry of the Lagrangian. To absorb this antisymmetry, we define a new variable: \[ \Phi = -\Phi_\#. \] This ensures: \[ \bar{\mathcal{L}}(\dot{\Phi}, \Phi) = \frac{C\dot{\Phi}^2}{2} - \frac{\Phi^2}{2L}. \]

The corresponding canonical momentum \(\Pi\) (related to \(Q\) based on previous results): \[ \Pi \equiv \frac{\partial \bar{\mathcal{L}}}{\partial \dot{\Phi}} = C\dot{\Phi} = -C\dot{\Phi}_\# = Q \implies \dot{\Phi} = \frac{\Pi}{C} = \frac{Q}{C}. \]

The Hamiltonian \(\bar{\mathcal{H}} \) is defined as: \[ \bar{\mathcal{H}} = \Pi\dot{\Phi} - \bar{\mathcal{L}} = \frac{\Pi^2}{C} - \frac{\Pi^2}{2C} + \frac{\Phi^2}{2L} = \frac{\Pi^2}{2C} + \frac{\Phi^2}{2L}, \] which matches the previous \(\mathcal{H}\) exactly. The quantization condition becomes: \[ [\Phi, \Pi] = [\Phi, Q] = i\hbar, \] identical to the traditional condition.

Define new variables \(a\) and \(a^\dagger\): \[ a = \sqrt{\frac{1}{2\hbar} \sqrt{\frac{L}{C}}} \left( \sqrt{\frac{C}{L}} \Phi + iQ \right), \quad a^\dagger = \sqrt{\frac{1}{2\hbar} \sqrt{\frac{L}{C}}} \left( \sqrt{\frac{C}{L}} \Phi - iQ \right). \] Conversely: \[ \Phi = \sqrt{\frac{\hbar}{2} \sqrt{\frac{L}{C}}} (a^\dagger + a), \quad Q = i\sqrt{\frac{\hbar}{2} \sqrt{\frac{C}{L}}} (a^\dagger - a). \]

It is easy to verify \([a, a^\dagger] = 1\): \[ [a, a^\dagger] = aa^\dagger - a^\dagger a = \frac{1}{2\hbar} \sqrt{\frac{L}{C}} \left[ \Phi, Q \right] = 1. \] The Hamiltonian \(H(\Phi, Q)\) can be rewritten as: \[ \mathcal{H} = \frac{\Phi^2}{2L} + \frac{Q^2}{2C} = \frac{\hbar}{4\sqrt{LC}} \left[ (a^\dagger + a)^2 - (a^\dagger - a)^2 \right]. \] Given \(\omega = \frac{1}{\sqrt{LC}}\), expanding this yields: \[ \mathcal{H} = \frac{\hbar \omega}{4} \left[ (a^\dagger + a)^2 - (a^\dagger - a)^2 \right] = \hbar \omega \left[ a^\dagger a + \frac{1}{2} \right]. \]

Finally, we once again obtain the quantized Hamiltonian, where the difference between each energy level is \(\hbar \omega\).

Originally written in Chinese by the author, these articles are translated into English to invite cross-language resonance.